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I have a function in C that should return the most negative two's complement number:

int mostNegTwosComp(void) {

    return 0;
}

I am constrained to using a maximum of 4 bitwise operators. These operators include: ! ~ & ^ | + << >>. How would I go about doing this? Wouldn't the most negative two's comp number be dependent on how many bits the selected number is? For instance, 10000 would be the most negative two's comp number of a 16 bit int?

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  • ! and + are not bitwise operators. Are we to take you to mean simply that the list you've given comprises all the operators you are permitted to use? Commented Feb 5, 2019 at 1:07
  • @EricPostpischil I'm honestly not entirely clear on what the question means, but I really didn't state my confusion well. I still can't, and it seems enough others do agree on a meaning, so deleted the comment. Commented Feb 5, 2019 at 1:34
  • @aschepler: The questions asks to produce the int value that is most negative (has the minimum value of all int values) using at most four of the operators listed. Commented Feb 5, 2019 at 1:52
  • you are looking for (signed)(~(~0U>>1)) But beware that this implies your internal representation is two's complement. Otherwise, you'll get U.B. Commented Feb 6, 2019 at 18:21

2 Answers 2

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If you don't need a portable version, you can abuse the knowledge that integers are almost always 4 bytes. 

return 0x80000000;

In fact, if you know the size of the type you're going to return, you can skip the bitwise game and cheat with the format:

  1. In 0x__, each number is 4 bits.This means 2 digits are one byte.
  2. You want the first bit to be 1, and all other bits to be 0.
  3. 0x8 = 0b1000
  4. As such, you can represent the value as 0x80 + 2 '0's for every byte of the type past the first.

But to answer the rest of your question.

How would I go about doing this?

If you're templating, you'd (probably) use the bitwise trick the other answer suggests. Otherwise you can cheat with the above code or use a definition from limits.h (iirc). 

~ (~0u >> 1);

Would be a portable solution.

Wouldn't the most negative two's comp number be dependent on how many bits the selected number is?

The most negative two's compliment is dependent upon the size of the containing variable, so I suppose you could say "selected number". In fact, the range of values depends on the size of the containing variable.

For instance, 10000 would be the most negative two's comp number of a 16 bit int?

For 16 bits, the most negative two's comp would be 0x8000, 0b1000000000000000 or -32768, depending on how you'd like it represented.

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6 Comments

Although very likely to work in practice, ~ (~0u >> 1) is not a portable solution, for the reason that the other answer already gives, and also because it (and your other) assume two's-complement representation. But it is more portable than your lead suggestion, because it does not rely on any assumptions about the width of type int.
Well, is there any relatively common configuration that'd not result in ~ (~0u >> 1) not being portable enough for most reasonable intents and purposes? I'm aware of why it isn't perfectly portable, but not aware where it isn't portable enough.
As I said, "very likely to work in practice." But "portable" is not a relative adjective in my book. I pretty much equate it with the Standard's "strictly conforming." Either code is, or it isn't, and ~ (~0u >> 1) isn't.
You are right, @Daniel, I have used sloppy, inconsistent wording in attempting to summarize the observation that ~ (~0u >> 1) relies on fewer assumptions.
@Daniel Brilliant! Thank you my man, your clear explanation was very helpful.
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return ~ (~0u >> 1);

~ turns on all bits in the unsigned zero. Then >> 1 shifts right, resulting in the high bit becoming zero. Then ~ inverts all bits, producing one in the high bit and zero in the rest.

Then the return converts this to int. This has implementation-defined behavior, but class assignments of this sort generally presume a behavior suitable for the exercise.

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